Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

For \(g(u)=3 /(u-2)\), find and simplify \([g(x+h)-\) \(g(x)] / h\).

Step-by-Step Solution

Answer

The simplified expression is \( \frac{-3}{(x-2)(x+h-2)} \).

Step 1: Plug in the Values

Start by writing out the expression \( [g(x+h) - g(x)] / h \) using the function \( g(u) = \frac{3}{u-2} \). Replace \( u \) with \( x+h \) and \( x \) in \( g(u) \).Thus, \[g(x+h) = \frac{3}{(x+h)-2} = \frac{3}{x+h-2}\]\[g(x) = \frac{3}{x-2}\]So, the expression becomes \[\frac{3}{x+h-2} - \frac{3}{x-2}\] divided by \( h \).

Step 2: Simplify the Expression

To simplify \( \frac{3}{x+h-2} - \frac{3}{x-2} \), find a common denominator. The common denominator here is \[(x+h-2)(x-2).\]Rewrite each fraction:\[\frac{3}{x+h-2} = \frac{3(x-2)}{(x+h-2)(x-2)}\]\[\frac{3}{x-2} = \frac{3(x+h-2)}{(x-2)(x+h-2)}\]Now subtract these fractions:\[\frac{3(x-2) - 3(x+h-2)}{(x-2)(x+h-2)}\]

Step 3: Simplify the Numerator

Simplify the expression in the numerator:\[3(x-2) - 3(x+h-2) = 3x - 6 - (3x + 3h - 6)\]Distribute and combine terms:\[= 3x - 6 - 3x - 3h + 6 = -3h\]

Step 4: Divide by h

Now substitute back into the main expression:\[\frac{-3h}{(x-2)(x+h-2)} \times \frac{1}{h}\]This simplifies to \[\frac{-3}{(x-2)(x+h-2)}\] after canceling \( h \) in the numerator and denominator.

Key Concepts

Difference Quotient

The difference quotient is a critical concept in calculus, as it forms the basis for understanding derivatives, which describe how a function changes at any given point. In our exercise, we're dealing with the difference quotient expressed as \( \frac{g(x+h) - g(x)}{h} \). Let's break this down:

• First, substitute specific values into the function \( g(u) \). For our problem, this means replacing \( u \) with both \( x+h \) and \( x \) to get their respective functions.
• Then, calculate \( g(x+h) \) and \( g(x) \), leading to an expression \( \frac{3}{x+h-2} - \frac{3}{x-2} \).
• Finally, divide the difference by \( h \) to establish the difference quotient itself.

This strategy is all about understanding how the function behaves as it moves from one point to another nearby. It's a stepping stone towards determining the rate at which a function is changing, laying the groundwork for derivatives.

Function Simplification

Simplifying functions is a critical skill in calculus that involves making an expression easier to work with. This typically means reducing a complex fraction or expression to a more manageable form. Here's how we simplify the expression from the exercise:

• Identify the need for a common denominator to add or subtract fractions. For example, for the fractions \( \frac{3}{x+h-2} \) and \( \frac{3}{x-2} \), the common denominator becomes \((x+h-2)(x-2)\).
• Re-express each fraction with this common denominator, turning the expression into \( \frac{3(x-2) - 3(x+h-2)}{(x+h-2)(x-2)} \).
• Focus on simplifying the numerator by distributing and combining like terms, resulting in \(-3h\).

Always remember that simplifying makes evaluating expressions and understanding the function's behavior much easier, especially when approaching a calculus problem like ours.

Rational Functions

Rational functions are functions expressed as the quotient of two polynomials. The function \( g(u) = \frac{3}{u-2} \) from our exercise is a simple example. Here we explore some key points about rational functions:

• The denominator cannot be zero, so in \( g(u) \), \( u-2 \) cannot equal zero. This implies \( u \) cannot be 2, helping identify potential points of discontinuity.
• These functions often require simplification techniques, involving operations like addition and subtraction of fractions, always demanding a common denominator.
• Simplification processes lead to clearer interpretations of the function's behavior, especially when calculating values around problematic points (like vertical asymptotes).

Understanding rational functions and their behavior is essential as it prepares you for tackling more complex calculus tasks, like limits and derivatives.